Sql 按以前的日期值分组
我正在尝试获取每个组的最小行和最大行,其结束日期与下一行开始日期匹配 输入日期:Sql 按以前的日期值分组,sql,sql-server,sql-server-2012,Sql,Sql Server,Sql Server 2012,我正在尝试获取每个组的最小行和最大行,其结束日期与下一行开始日期匹配 输入日期: ResultUid BeginDate EndDate 1 1999-12-31 00:00:00.000 2000-01-31 00:00:00.000 1 2000-01-31 00:00:00.000 2000-02-29 00:00:00.000 1 2000-02-29 00:00:00.000 2
ResultUid BeginDate EndDate
1 1999-12-31 00:00:00.000 2000-01-31 00:00:00.000
1 2000-01-31 00:00:00.000 2000-02-29 00:00:00.000
1 2000-02-29 00:00:00.000 2000-03-31 00:00:00.000
1 2000-03-31 00:00:00.000 2000-04-30 00:00:00.000
1 2007-03-31 00:00:00.000 2007-04-30 00:00:00.000
1 2007-04-30 00:00:00.000 2007-05-31 00:00:00.000
1 2007-05-31 00:00:00.000 2007-06-30 00:00:00.000
预期结果:
ResultUid BeginDate EndDate
1 1999-12-31 00:00:00.000 2000-04-30 00:00:00.000
1 2007-03-31 00:00:00.000 2007-06-30 00:00:00.000
我试过:
SELECT
ResultUid,
MIN(BeginDate) AS "min",
MAX(EndDate) AS "max",
lag
FROM (
SELECT
ResultUid,
BeginDate,
EndDate,
DATEDIFF(MONTH,lag(BeginDate) OVER (order by EndDate), EndDate) AS "lag"
FROM Results
GROUP BY
ResultUid,
BeginDate,
EndDate
) sub
GROUP BY
ResultUid,
lag
您可以通过检查上一个结束日期来确定组的起始位置。然后,相邻日期的组可以通过采用累积总和来分配组id:
select resultuid, min(begindate) as begindate, max(enddate) as enddate
from (select r.*,
sum(case when prev_enddate = begindate then 0 else 1 end) over
(partition by resultuid order by begindate) as grp
from (select r.*,
lag(enddate) over (partition by resultuid order by begindate) as prev_enddate
from results r
) r
) r
group by resultuid, grp;